Item 9 gull eggs 4 ppm Lake trout 4.83 ppm melt 1.04 ppm zooplankton Phytoplankt
ID: 165759 • Letter: I
Question
Item 9 gull eggs 4 ppm Lake trout 4.83 ppm melt 1.04 ppm zooplankton Phytoplankton 0.123 ppm 0.025 ppm Pearson Education, Inc. Biological magnification of PCBs has been found in the food wab af the Great Lakes, where the concentration of PCBs in herring gull eggs, at the top of the food web. is nearly 5,000 times that in phytoplankton, at the base of the food web. In the figure, ppm E parts per million. If you need a review of scientific notation, try this practice exercise. Part A If a typical smelt weighs 225 g, what is the total mass of PCBs in a smelt in the Great Lakes? 2.34 x 10 4 g 225 g 234 g 2.16 x 108 g Submit My Answers Part B lf an average lake trout weighs 4,500 g, what is the total mass of PCBs in a trout in the Great Lakes? 2 17 x 10 g 4,500 g 21,700 g 932 x 103 g My Answers Give Up SubmiExplanation / Answer
Ans. Part A. Given, [PCBs] in smelt = 1.04 ppm (see picture to the right).
1.04 ppm means 1.04 mg per kg.
That is, 1.04 ppm = 1.04 mg/ kg body mass = 1.04 mg/ 1000 g body mass
That is, a smelt weighing 1.000 kg (= 1000.0 g) will have 1.04 mg in its body.
Given, mass of smelt = 225 g
Now,
Total mass of PCBs in smelt = Mass of smelt x concentration of PCBs in it
= 225 g x (1.04 mg/ 1000 g)
= 0.234 mg
= 0.000234 g ; [1 mg = 0.001 g]
= 2.34 x 10-4 g
Thus, correct option. A
Part B. Given, [PCBs] in lake trout = 4.83 ppm (see picture to the right). That is, a lake trout weighing 1.000 kg (= 1000.0 g) will have 4.83 mg in its body.
Given, mass of lake trout = 4500 g
Now,
Total mass of PCBs in lake trout = Mass of lake trout x concentration of PCBs in it
= 4500 g x (4.83 mg/ 1000 g)
= 21.735 mg
= 0.021735 g [1 mg = 0.001 g]
= 2.17 x 10-2 g
Thus, correct option. A
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